Proof for the problem x^5 - y^2 = 4

[ejiant]

I may have proof here.

-y^2 = 4 - x^5
y^2 = x^5 - 4
y = sqrt(x^5 - 4)
sqrt(x^5 - 4) must be an integer.

In order for a number to have an integer square root, each of prime factors must have an even number (i.e. 2^2 * 3^2 * 5^6) . That is because the primes must be able to be split in half.

So I don't have to keep on typing x^5, let z = x^5.

Let's assume that sqrt(z) is an integer.

Therefore, if sqrt(z-4) were to still to have an integer square root, the subtraction of 4 must have taken out two prime numbers, p. // Edit: I'm not sure, but this logic might not be sound.

Edit 2: Nope. Drats. There goes my proof.

Edit 3: Well, I think I may have another solution.
z and z-4 must both be perfect squares.
1, 4, 9, 25, etc. are perfect squares.
The differences between these squares is an arithmetic series with a cd of 2, starting at 3. Therefore, z and z-4 can never both be perfect squares.


Ignore
-----------------

z-4 = z/(p^2)
z = z/(p^2) + 4
zp^2 = z + 4p^2
zp^2 - 4p^2 - z = 0
(z-4)p^2 - z = 0
Therefore, it is a quadratic equation where a = (z-4), b = 0, and c = -z

In order for the quadratic equation to have a real solution, sqrt(b^2 - 4*a*c) must be >= 0.

sqrt(b^2 - 4*a*c) here is sqrt(0 - 4*(z-4)(-z))
sqrt(0 - (4z-16)(-z))
sqrt(0 - (-4z^2 + 16z))
sqrt(4z^2 - 16z)
If 4z^2 - 16z must be greater or equal to 0,
4z^2 >= 16z.
z^2 >= 4z
z >= 4

The whole quadratic equation for (z-4)p^2 - z would be
0 + (can't be minus) sqrt(4z^2 - 16z)/(2*(z-4))
sqrt(4z^2 - 16z)/(2z-8)
Square both the numerator and denominator:
(4z^2 - 16z)/(4z^2 - 32z + 64)
4z^2 - 16z > 4z^2 - 32z + 64
-16z > -32z + 64
-16z > 64 - 32z
16z > 32z - 64
0 > 16z - 64
64 > 16z
4 > z

Proof by contradiction.


I'm a freshman in highschool, so please forgive any problems I have with presenting a proof.

I also hope I didn't make a silly arithmetic error, screwing my whole proof up.

[yoyo]

The only residues of x^5 (mod 11) are 0, 1, -1.

The quadratic residues modulo 11, or the possible values of y^2 (mod 11), are 0, 1, 4, 9, 5, 3.
So the possible values of y^2 + 4 (mod 11) are 4, 5, 8, 2, 9, 7.

So x^5 cannot equal y^2 + 4 (mod 11), and x^5 - y^2 = 4 (mod 11) has no integer solutions. Hence, x^5 - y^2 = 4 has no integer solutions.


I found 11 through a brute force search. Is there a general way to figure out what you can mod x^n by so that the only values are {0, 1, -1}?

[rrusczyk]

Why must z-4 be z/p^2?

It might be some perfect square that has factors other than those of z. For example, suppose 4 were 40 instead: 7^2-40=3^2.

[ejiant]

Right. I realize that both of my solutions are wrong.

[rrusczyk]

Nice solutions, yoyo.

How did you think to use mods?

Are there any budding number theorists out there who want to take a swing at yoyo's general question?

[i/3]

If p > 2 is prime, a^(p-1) = 1 (mod p) if (a,p) = 1 (Fermat little theorem)

So (a^(p-1)/2)^2 = 1 (mod p), so (Z/pZ is a field) a^(p-1)/2 in {1,-1}


Conclusion :

if p is prime > 2, the values of a^(p-1)/2 are in {0, 1, -1}


Set p =11

[yoyo]

[akatookey]

I am interested in yoyo's (Tim's?) question...is there a way to find a mod n where the solutions of x^n are only {-1,0,1} ???

[rrusczyk]

[TripleM]

[TripleM]

Heres another problem with exactly the same idea as the above one. Prove that the equation x^3 + y^3 + z^3 = 2003 has no integer solutions. (Hint : think mods!!)

[i/3]

2003 mod 9 = 5

x^3 mod 9 in {0,1,-1} so x^3 + y^3 + z^3 always different from 5 mod 9.


Why 9 ???


Euler theorem :

a^phi(n) = 1 mod n if (a,n) = 1 (n prime or not).
phi being the Euler function

phi(9) = 6

so a^6 = (a^3)^2 = 1 if (a,9) = 1.

[andy17null]

I don't have anything definite yet, but I was messing around with this in math class. Since sqrt(x^5-4) has to be an integer, you need to find an integer^5 that is 4 greater than a square. I tested this for x'es of 2 to 10 and this is what I got...
X----x^5-4---Nearest square---------Value of-------Difference between
------------number below x^5-4----that square----x^5-4 and the square
2___28________5^2_________________25_______________3
3___239_______15^2________________225_____________14
4___1020______31^2________________961_____________59
5___3121______55^2________________3025____________96
6___7772______88^2________________7749____________23
7___16803_____129^2_______________16641__________162
8___32764_____181^2_______________32761__________3
9___59045_____242^2_______________58564__________481
10__99996_____316^2______________99956___________140

I tested this for only 1,2,3, and 4 at first. I was really excited because I noticed that the differences kept on increasing. That theory was blown away when I continued the pattern. I put this up here because someone might be able to use it, perhaps there is a pattern in the differences or the squares?
Sorry about the messy underscores- the posting program deletes additional spaces. Hope that you can read this okay.


-Isaac